Stoichiometry of chlorine reactions with different reagents

Steps:

• Calculate moles of Cl₂: 0.710 g ÷ 71 g/mol = 0.010 mol Cl₂.
• First reaction (with H₂): Cl₂ + H₂ → 2HCl; 0.020 mol HCl × 36.5 g/mol = 0.730 g HCl.
• Second reaction (with NaOH(aq)): Cl₂ + 2NaOH → NaCl + NaOCl + H₂O; 0.010 mol NaCl × 58.5 g/mol = 0.585 g NaCl.
• Third reaction (with hot NaOH(aq)): Same hypochlorite equation as second (no concentration specified for chlorate); 0.585 g NaCl.

Why A is correct:

• Balanced equation Cl₂ + H₂ → 2HCl yields 2 mol HCl per mol Cl₂, so 0.730 g matches the stoichiometry.

Why the others are wrong:

• B: Second reaction yields 0.585 g NaCl per balanced equation, not 0.858 g.
• C: Third reaction (dilute aqueous, even hot) yields 0.585 g NaCl, not 0.975 g (distractor for chlorate equation).

Final answer: A